Simple dipstick problem

[pat666]

1. The problem statement, all variables and given/known data

I have a problem involving a dipstick, a cone and a cylinder. Basically I want to make a dipstick for a cone and cylinder which shows the volume of remaining fluid in the containers.

2. Relevant equations



3. The attempt at a solution
for the cone
V=pi r^2 *h/3
now I cant find how to relate depth to volume?

please help

[pat666]

I just remembered to say that what I'm really really having a problem with is finding the correct relationship between height and radius.

Thanks

[ebits21]

Well, you could start by focusing on the cylinder.

If the volume of the cylinder is volume = pi * radius^2 * height

Can you rewrite the equation so that you solve for height? If you can do that you know that a certain height on the dipstick will correspond to a certain volume of the cylinder.

Then you can figure out how the cone will add on to that.

[pat666]

the cylinder is on its side so its not that simple I don't think??? sorry I forgot to say that in the OP.

[ebits21]

In that case you need to rewrite the volume equation for the cylinder in terms of the diameter of the cylinder.

Remember that radius = 1/2 the diameter.

That way you'll know the volume that corresponds to the "diameter height" that you have written on the dipstick.

Is the cone attached to the cylinder or is it a separate question?

[pat666]

the cone is a separate question. so with the diametre V=pi*1/4 D^2 *h then D=sqrt(4V/pi) but that will only work until 1/2 way won't it?

[ebits21]

You know what, I think I actually made a mistake. I think you're actually going to have to use some trigonometry to solve this.

I'll have to think about this a bit.

[pat666]

thanks Ill wait for your reply

[ebits21]

Well, I can say this much. If you look at the cylindrical tank from one of the ends, you would have a circle with the liquid coming up a certain height of the circle.

Therefore the problem becomes a question of finding the area of the segment of the circle below the chord marked by the top of the liquid. This is then multiplied by the length of the cylinder to find the volume.

Since it's a circle, you can draw a line from the center of the circle to the ends of the chord that marks the height of the liquid. Bisecting this, you get two right angle triangles with a hypotenuse equal to the radius, and a height equal to the radius minus the height of the liquid. The angle between these two is then used to figure things out...

Unfortunately... I'm kind of rusty on this so I'm afraid I can't be of much more help. Maybe someone else can take it from here?

[pat666]

What you are saying at the start makes perfect sense to me and hopefully someone can help me further with the math.

Thanks

[ebits21]

I thought I would mention another thing.

If you take the area of the "pie slice" that is marked by the lines drawn to the top of the chord (including the area that is swept out to the bottom of the tank), and then you subtract the area of the triangles above the liquid, that should give you the area of the liquid in the circle.

[ebits21]

I found this on wikipedia... which might help http://en.wikipedia.org/wiki/Circular_segment

You need to rewrite the area formula in terms of the height of the liquid, and I think that R = h + (r-h) gives you a way to do that.

[Mentallic]

I'm willing to give this question a go, but I can't seem to concentrate at the moment, probably because of my headache. If anyone could give me a jump start with what we're trying to solve, I'll give a lending hand to finish the problem off.

[pat666]

Hey Mentallic, any/all help is appreciated. Were trying to make a dipstick for a sideways cylinder and a cone. Currently I'm having trouble relating the height of the liquid on the circle to the area of that part.

Thanks

[Mentallic]

Ok since you said in post #6 that the cylinder and cone are separate questions, I'm going to assume that you want a relation between the height and volume of liquid present in a cone on its side (like a wheel) and the cone will be laying down which way exactly?

First of all, do you know about and are you supposed to use integration?

[pat666]

Cone is upright, cylinder is on its side. I know integration but this is for an 11th grader I'm helping, I seem to have forgotten everything from then. No integration for him yet though. It is the cone question that I need help with as there is ample information on the cylinder on the net.

Thanks and hopefully the fact that its upright (ice cream cone ways) makes this simpler and sorry for not stating this directly.

[Mentallic]

Alright sure, and yes it does make the problem easier :biggrin:

Ok so I'm guessing that the dimensions of the cone are known constants. Let's give it a base radius R and perpendicular height H, so the volume of the cone is V=\frac{\pi R^2 H}{3}

Now let's fill this cone up a bit with water, the water level will be at a height h. Now, just look at the part of the cone that isn't filled with water - it's a smaller cone with height H-h. Let the radius of this cone be r.

So the volume of the water is now simply V=\frac{\pi R^2 H}{3}-\frac{\pi r^2 (H-h)}{3}

Ok so what constants do we know in this formula? We know R, H, h (because this will be observed on the dipstick) and we don't know r. Well r is pretty easy to find, just take a side-view of the cone and look at one side of the cone. It will be a right triangle with height H and length R. There will be another similar but smaller right triangle within it with height H-h and length r. Since these triangles are similar, their dimensions are proportional to each other. Can you take it from here?

[pat666]

I should be able to.One thing that I'm not totally sure about is the "cone" that isn't filled with water, is this a cone, or at least does its volume follow the standard volume formula for a cone??

[pat666]

also now that I'm thinking about it I also can't find r, sorry I just didn't realize how much of this I've forgotten. Fairly embarrassing that I can't do yr 11 math anymore!!!

Thanks

[Mentallic]

I should be able to.One thing that I'm not totally sure about is the "cone" that isn't filled with water, is this a cone, or at least does its volume follow the standard volume formula for a cone??
Of course it's a cone. It has the same apex (point) as the larger cone but a shorter height and thus shorter radius. The cones are similar and both have flat bottoms (because water settles due to gravity of course :biggrin:).

also now that I'm thinking about it I also can't find r, sorry I just didn't realize how much of this I've forgotten. Fairly embarrassing that I can't do yr 11 math anymore!!!

Thanks
So we have two similar right triangles, the larger one with height H and length R, and the other one with height H-h and length r. Since they're similar, the proportions of each of their sides are in the same ratio. For example, a circle of radius 1 is similar to a circle of radius 2 such that their radii are a ratio of 1:2 and their circumferences are also a ratio of 1:2.
This means you can create the equality \frac{R}{r}=\frac{H}{H-h}

[pat666]

Perhaps I miss communicated the orientation of the cone. The cone is inverted like an Icecream cone. When I was looking for the picture of the cone I came across this.
http://btssmath3e.blogspot.com/2009/03/math-riddle-1-cone-paper-cups.html

From what I can see the cone without water has 2 flat sides.

Thanks

[Mentallic]

Oh sorry :yuck: When I read "the right way up" I assumed the way a cone can sit down without falling over or rolling everywhere.

Even so, the problem isn't very different at all. This time you have the large cone with height H and radius R and the smaller cone with height h radius r. Again, you know H, h and R, but you don't know r. The volume of the liquid is obvious since it's just a cone, but you just need to express r in terms of H, h and R. I've already shown you the method of how you can do this.

[pat666]

This is the reason I couldn't do it last time, I can't see two similar triangle in this shape?

Also, since you have a good grasp of this would you mind checking what I've done with the horizontal cylinder: V=A (of end circle) *L
so the only thing I wanted to find out was the area of the circle that the liquid covered. which can be seen as 2 tight angle triangles between the midpoint and the liquad level.

so the Area of the liquid filled portion is A = pi*r^2/2 - r^2*arcsin(1-h/r) - (r-h)*sqrt(h(2r-h))

then that is just multiplied by length to get the volume
Thanks

[Mentallic]

http://img687.imageshack.us/img687/1209/coner.png (http://img687.imageshack.us/i/coner.png/)

Uploaded with ImageShack.us (http://imageshack.us)

[pat666]

So its the same as before essentially.
r/R=(H-h)/H

[Mentallic]

Sorry about the half-assed response before, I had to run.

No no, it is just \frac{H}{h}=\frac{R}{r} because the ratio between the big cone radius (R) to the little cone radius (r) is the same as the big cone height (H) to the little cone height (h). So you pretty much have that problem answered once you solve for r in terms of the other known constants and then plug that into the volume of the small cone formula.

I'm sure the formula for the cylinder can be a lot simpler than that, I'll solve the problem now and then get back to you.

[pat666]

No worrys, You've been way more than helpful so far. THANKS
r=(h/H)*R

V=(\pi*R^2*h)/3 -\pi*((h/H)*R)^2(H-h)/3 hope I've done the LaTex correctly.

The way I read that is V entire cone minus V water filled portion of cone and what that yields is the volume of the empty portion of the cone. So my question is should That answer now be subtracted from the volume of the cone again??

Thanks

[Mentallic]

so the Area of the liquid filled portion is A = pi*r^2/2 - r^2*arcsin(1-h/r) - (r-h)*sqrt(h(2r-h))

Can you show me how you got that, because I'm getting something else.

[Mentallic]

No worrys, You've been way more than helpful so far. THANKS
r=(h/H)*R

V=(\pi*R^2*h)/3 -\pi*((h/H)*R)^2(H-h)/3 hope I've done the LaTex correctly.
Don't put spaces in the and [/tex ] tags and then you'll get latex code :wink:
edit: never mind, you fixed it already.

The way I read that is V entire cone minus V water filled portion of cone and what that yields is the volume of the empty portion of the cone. So my question is should That answer now be subtracted from the volume of the cone again??

Thanks

Sure, you can do it that way, but why not take the much easier approach? You already know the volume of the smaller water-filled cone is [tex]V=\frac{\pi r^2 h}{3} and you don't know r, but you have already shown what r is in terms of R, H and h. Just plug it into there and you're set. You should also get the same answer from doing it the longer way as you've suggested.

[pat666]

Theres every chance that I'm wrong, as I said I am very rusty at this. Perhaps this picture will help illustrate what I'm thinking. As I'm looking at the pic I just drew I think I may have stuffed something up, its not making sense to me now.

[pat666]

oh yeah, don't know why I missed that.
so V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.

Thanks

[Mentallic]

Theres every chance that I'm wrong

Haha fair enough :biggrin:

Since we've already failed a few times at understanding each other with much simpler diagrams, I'll have to draw up a picture for this more complicated one.

First of all, we can save ourselves a lot of time if we don't have to prove the area of the segment of a circle. The wiki article for it was already posted:
I found this on wikipedia... which might help http://en.wikipedia.org/wiki/Circular_segment

If you don't understand it though just ask and I'll walk you through it.

Ok so we have that the area of the segment of the circle is A=\frac{r^2}{2}\left(\theta-\sin\theta\right)

But we don't know \theta so again we're going to have to do something like in the other question to find \theta in terms of other constants that we do know, mainly h (the height of the water level) and r (the radius of the circle).

http://img17.imageshack.us/img17/1326/circleo.png

The angle \theta sweeps the entire brown angle in the circle picture, and therefore due to symmetry, the brown angle in the triangle picture is going to be \frac{\theta}{2}

Now you can easily get your relationship and plug it back into the equation. And by the way, \sin\left(\cos^{-1}\left(x\right)\right)=\sqrt{1-x^2}

[Mentallic]

V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.

Sure, V=\frac{\pi h^3R}{3H^2}

By the way, you can click on my LaTeX to see how it is written, in case you're interested.

[pat666]

I get the circular segment part (to some degree), what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent or is this a misprint or an error on my part? Also just to confirm, we are simply trying to get theta in terms of the depth of the water?

P.S just to confirm the last post I made on the cone question is correct?
Thanks

[Mentallic]

what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent
Why does r need to equal h? r is the radius of the circle and h is the height of the water. First let h<r, so that the side of the triangle is r-h because it is what is left of the radius after taking away h. For r=h we have a degenerate case where the water fills up half the circle, so it would simply be A=\frac{\pi r^2}{2}. You can also take h> r and try find the formula for that, but it'll be the same as for h<r.

Also just to confirm, we are simply trying to get theta in terms of the depth of the water?
In terms of the depth of the water AND the radius of the circle. We can't do it in terms of the depth of the water alone. Imagine we had a circle radius 1 and the depth of the water was very close to it, about 0.99. The angle \theta will be very close to \pi . Now what if we kept the water depth the same, but made the circle very big now? Well the angle will get a lot smaller. So obviously \theta is dependent on more than just the depth of the water.

P.S just to confirm the last post I made on the cone question is correct?
Thanks
Well you tell me, what makes you unsure if it is correct or not?

[pat666]

I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure. just the way the picture was drawn made me think r=h, which would be true if the water was at the centre. I get that \theta=2cos^-^1(r-h)/r
not sure if thats correct because you gave me some trig info that was a bit more complex.
also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks

[Mentallic]

I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure.
Yes it's correct. I was just showing you that you can tidy up
V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.
into
Sure, V=\frac{\pi h^3R}{3H^2}


just the way the picture was drawn made me think r=h,
Oh, yeah that would be my fault, sorry :smile:

which would be true if the water was at the centre.
The blue line was meant to be where the water level was at.

I get that \theta=2cos^-^1(r-h)/r
Yes that's right.

not sure if thats correct because you gave me some trig info that was a bit more complex.
My trig info was wrong, ignore it. I forgot about the 2 that was going to be in front of it. What I was meant to give you was
\sin\left(2\cos^{-1}\left(x\right)\right)=2x\sqrt{1-x^2}
It may look complex, but its purpose is simple. When you plug \theta into the area equation A=\frac{r^2}{2}\left(\theta-\sin\theta\right) you're going to be left with
edit: \sin\left(2\cos^{-1}\left(\frac{r-h}{r}\right)\right) which is where you can simplify this with the equality I gave above.
You already have the answer, but it's just if you wanted to simplify things a bit more.

Markers wouldn't give full marks if you left an answer as \sin\left(\sin^{-1}\left(x\right)\right) so I doubt they would give full marks if you left it as \sin\left(\cos^{-1}\left(x\right)\right) either.

also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks
I believe the same formula will work for the water level anywhere from 0 to 2r (the diameter of the circle) but I'll check to see.

[pat666]

Ok thanks alot for all your help

[OmCheeto]

The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.
There is a reason they put this problem in computer science text books and never in mathematics text books.

[Mentallic]

Ok thanks alot for all your help
No worries

The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.
Can you please elaborate?

There is a reason they put this problem in computer science text books and never in mathematics text books.
I've seen questions similar to this in maths books.

[pat666]

This question is for yr 11 math. My solution is I believe V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right) .

[Mentallic]

This question is for yr 11 math. My solution is I believe V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right) .

That's not right. The formula is A=\frac{r^2}{2}\left(\theta-\sin\theta\right) where \theta=2\cos^{-1}\left(\frac{r-h}{r}\right)

[pat666]

whoops I forgot the first theta, so V=L*(r^2/2(2arccos(r-h/r)-sin(2arccos(r-h/r))) pretty messy.

[OmCheeto]

Can you please elaborate?

Perhaps the question in the book was worded differently.
All I know is that if you have a 10 gallon tank, it is not possible to place integer gallon marks on the dipstick. Unless of course you are clever enough to solve for x.


And perhaps I should give some background (http://www.physicsforums.com/showthread.php?p=3042826#post3042826) regarding this particular problem before someone yells at me for playing mind games.